c - 指向 void 指针的指针需要显式转换

Question

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Answer 1

Chapter 22: Pointers to Pointers :

One side point about pointers to pointers and memory allocation: although the void * type, as returned by malloc, is a "generic pointer," suitable for assigning to or from pointers of any type, the hypothetical type void ** is not a "generic pointer to pointer".

因此，只有 void * 是通用指针。 void ** 不是通用指针。因此，传递给函数的参数必须是 void ** 类型。

另见 C-FAQ :

There is no generic pointer-to-pointer type in C. void * acts as a generic pointer only because conversions (if necessary) are applied automatically when other pointer types are assigned to and from void *'s; these conversions cannot be performed if an attempt is made to indirect upon a void ** value which points at a pointer type other than void *. When you make use of a void ** pointer value (for instance, when you use the * operator to access the void * value to which the void ** points), the compiler has no way of knowing whether that void * value was once converted from some other pointer type. It must assume that it is nothing more than a void *; it cannot perform any implicit conversions.

In other words, any void ** value you play with must be the address of an actual void * value somewhere; casts like (void **)&dp, though they may shut the compiler up, are nonportable (and may not even do what you want; see also question 13.9). If the pointer that the void ** points to is not a void *, and if it has a different size or representation than a void *, then the compiler isn't going to be able to access it correctly.